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SS2 PHYSICS SECOND TERM SCHEME OF WORK
Week 1: Coordinate systems: position & distance
Week 2: Scalars & vectors
Week 3: Equations of uniformly accelerated motion
Week 4: Projectiles
Week 5: Introduction to Simple Harmonic Motion (SHM)
Week 6: Applications of SHM (pendulum, springs, etc.)
Week 7: Force and equilibrium of particles
Week 8: Equilibrium of rigid bodies & moments of forces
Week 9: Center of gravity, stability & equilibrium in fluids (Archimedes’ principle, flotation)
Week 10: Revision
Week 11: Examination
Week 12: Closing
WEEK 1 LESSON NOTE
Subject: Physics
Class: Senior Secondary School 2 (SS2)
Term: Second Term
Week: One (1)
Topic: Coordinate Systems: Position and Distance
Duration: 40 minutes × 2 periods
SPECIFIC OBJECTIVES
By the end of this lesson, students should be able to:
Explain the meaning of a coordinate system in physics.
Identify and describe different types of coordinate systems used in physics.
Define position, distance, and displacement.
Represent the position of an object using a coordinate system.
Calculate distance between two points on a straight line using coordinates.
Distinguish clearly between distance and displacement with examples.
Apply coordinate systems to simple real-life physical situations.
INSTRUCTIONAL MATERIALS
Graph sheets
Ruler
Meter rule
Physics textbook
Charts showing Cartesian coordinate axes
Real-life objects (ball, book, box)
LESSON CONTENT
The lesson is organized in clear steps, with each step addressing one or more of the stated objectives.
STEP 1: MEANING OF A COORDINATE SYSTEM
A coordinate system is a method used to describe the position of a point or object in space using numbers called coordinates.
In physics, coordinate systems help us:
Describe the exact position of an object
Measure distance and displacement
Study motion in one, two, or three dimensions
Without a coordinate system, it would be difficult to explain where an object is located or how it moves.
Example:
Saying “the book is over there” is vague, but saying “the book is 2 m to the right of the table” is precise.
STEP 2: TYPES OF COORDINATE SYSTEMS
Although several coordinate systems exist in advanced physics, at SS2 level, emphasis is placed on the Cartesian coordinate system.
(a) Cartesian Coordinate System
The Cartesian coordinate system consists of:
The point where the two axes meet is called the origin and is represented as (0, 0).
The axes divide the plane into four regions called quadrants:
First quadrant: positive x, positive y
Second quadrant: negative x, positive y
Third quadrant: negative x, negative y
Fourth quadrant: positive x, negative y
Each point on the plane is represented by an ordered pair (x, y), where:
STEP 3: MEANING OF POSITION
Position refers to the location of an object relative to a chosen reference point.
In physics, position is not absolute; it depends on:
Example:
If a tree is 10 m east of a house, the house can also be described as 10 m west of the tree. The position depends on the reference point.
In one-dimensional motion (straight line), position is often represented using only the x-axis.
STEP 4: REPRESENTING POSITION USING COORDINATES
To represent position using coordinates:
Choose a reference point (origin)
Choose a direction as positive
Measure the distance of the object from the origin
Example:
If a point A is 4 units to the right of the origin, its coordinate is (+4, 0).
If a point B is 3 units to the left of the origin, its coordinate is (-3, 0).
On a two-dimensional plane, a point C located 2 units right and 5 units up is written as (2, 5).
STEP 5: MEANING OF DISTANCE
Distance is defined as the total length of the path traveled by an object, irrespective of direction.
Characteristics of distance:
Unit of distance: metre (m)
Example:
If a student walks 5 m east and then 5 m west, the total distance covered is:
Distance = 5 m + 5 m = 10 m
STEP 6: DISTANCE ON A STRAIGHT LINE USING COORDINATES
When motion is along a straight line, distance between two points can be calculated using their coordinates.
If the coordinates of two points are x1 and x2, then:
Distance = |x2 − x1|
The vertical bars indicate absolute value, meaning the result is always positive.
Example:
If point A is at x = 2 m and point B is at x = 8 m:
Distance = |8 − 2| = 6 m
If point C is at x = -4 m and point D is at x = 3 m:
Distance = |3 − (-4)| = |7| = 7 m
STEP 7: MEANING OF DISPLACEMENT
Displacement is defined as the change in position of an object in a particular direction.
Displacement = Final position − Initial position
Characteristics of displacement:
It is a vector quantity (has magnitude and direction)
It can be positive, negative, or zero
It does not depend on the path taken, only on the initial and final positions
Unit of displacement: metre (m)
STEP 8: DIFFERENCE BETWEEN DISTANCE AND DISPLACEMENT
| Distance | Displacement |
|---|
| Scalar quantity | Vector quantity |
| Depends on path | Depends only on initial and final position |
| Always positive | Can be positive, negative, or zero |
| Measures total path | Measures shortest straight-line change |
Illustration:
If a boy walks from point A to B and returns to A:
Distance is not zero
Displacement is zero
STEP 9: REAL-LIFE APPLICATIONS OF COORDINATE SYSTEMS
Coordinate systems are used in many real-life situations, including:
Locating places on a map using latitude and longitude
Navigation by pilots and sailors
Position tracking in GPS systems
Motion analysis in physics and engineering
EVALUATION
The teacher asks the students the following questions:
What is a coordinate system?
Define position in physics.
State two differences between distance and displacement.
Calculate the distance between points at x = -2 m and x = 6 m.
Explain why displacement can be zero even when distance is not zero.
CLASS WORK
Draw a Cartesian coordinate plane and label the axes.
Plot the points (3, 2), (-2, 4), (-3, -1), and (4, -2).
Find the distance between x = 1 m and x = 9 m.
HOME WORK
Define coordinate system in your own words.
Explain distance and displacement with two examples each.
A man moves from x = -5 m to x = 7 m. Calculate:
(a) Distance traveled
(b) Displacement
CONCLUSION
The teacher summarizes the lesson by emphasizing that coordinate systems provide a clear and accurate way to describe the position and motion of objects in physics. Students are reminded that understanding position, distance, and displacement is fundamental to later topics such as motion, velocity, acceleration, and forces. The lesson is concluded with encouragement for students to practice drawing coordinate axes and solving simple numerical problems.
WEEK 2 LESSON NOTE
Subject: Physics
Class: Senior Secondary School 2 (SS2)
Term: Second Term
Week: Two (2)
Topic: Scalars and Vectors
Duration: 40 minutes × 2 periods
SPECIFIC OBJECTIVES
By the end of this lesson, students should be able to:
Define scalar and vector quantities.
Distinguish clearly between scalar and vector quantities.
Give examples of scalar and vector quantities in physics.
Explain the characteristics of vectors (magnitude and direction).
Represent vectors graphically.
Perform simple vector addition using graphical methods.
Apply the concept of scalars and vectors to real-life situations.
INSTRUCTIONAL MATERIALS
Graph sheets
Ruler
Meter rule
Physics textbook
Charts showing vector representation
Everyday objects for illustration (ball, stone, toy car)
LESSON CONTENT
STEP 1: MEANING OF SCALAR QUANTITIES
A scalar quantity is a physical quantity that has magnitude only and no direction.
Magnitude simply means the numerical value together with its unit.
Scalar quantities are completely described once their size or amount is stated.
Examples of scalar quantities include:
Mass (kg)
Time (s)
Temperature (°C or K)
Distance (m)
Speed (m/s)
Energy (J)
Volume (m³)
Explanation:
If a student says the distance covered is 10 m, the information is complete. There is no need to specify direction.
STEP 2: MEANING OF VECTOR QUANTITIES
A vector quantity is a physical quantity that has both magnitude and direction.
For a vector quantity to be fully described, both how much and the direction must be stated.
Examples of vector quantities include:
Displacement (m)
Velocity (m/s)
Acceleration (m/s²)
Force (N)
Momentum (kg m/s)
Weight (N)
Explanation:
Saying a body moves at 10 m/s is incomplete. The direction (e.g. eastward) must be stated.
STEP 3: DIFFERENCES BETWEEN SCALARS AND VECTORS
| Scalar Quantities | Vector Quantities |
|---|
| Have magnitude only | Have magnitude and direction |
| No direction involved | Direction is essential |
| Added algebraically | Added using vector laws |
| Examples: mass, time | Examples: force, velocity |
STEP 4: CHARACTERISTICS OF A VECTOR
Every vector quantity has two main characteristics:
Magnitude – the size or numerical value of the quantity.
Direction – the line along which the quantity acts and the sense of action.
A vector is usually represented by a straight line with an arrow head.
STEP 5: REPRESENTATION OF VECTORS
Vectors can be represented in different ways:
(a) Symbolic Representation
Vectors are represented using bold letters or arrow signs.
Examples:
(b) Graphical Representation
To represent a vector graphically:
Choose a suitable scale (e.g. 1 cm = 2 N).
Draw a straight line according to the magnitude.
Add an arrow head to show direction.
STEP 6: ADDITION OF VECTORS (GRAPHICAL METHOD)
Vectors are added using graphical methods, especially when direction is involved.
(a) Head-to-Tail Method
Steps:
Draw the first vector to scale.
From the head of the first vector, draw the second vector.
The resultant vector is drawn from the tail of the first vector to the head of the second vector.
(b) Parallelogram Method
Steps:
Draw the two vectors from the same point.
Complete the parallelogram.
The diagonal of the parallelogram represents the resultant vector.
STEP 7: REAL-LIFE APPLICATIONS OF SCALARS AND VECTORS
Speed limit signs show scalar quantities.
Navigation of ships and aircraft uses vectors.
Forces acting on structures are vector quantities.
Wind velocity is described using magnitude and direction.
EVALUATION
Define scalar quantity.
Define vector quantity.
Give five examples each of scalar and vector quantities.
State two differences between scalars and vectors.
Explain how vectors are represented graphically.
CLASS WORK
Classify the following as scalar or vector: speed, velocity, mass, force, time.
Draw a vector of magnitude 6 units using a suitable scale.
Explain why displacement is a vector quantity.
HOME WORK
List ten scalar quantities.
List ten vector quantities.
Explain with a diagram how two vectors can be added using the head-to-tail method.
CONCLUSION
The lesson is concluded by emphasizing that understanding scalars and vectors is essential in physics because many physical quantities involve both magnitude and direction. Students are reminded that this topic forms the foundation for studying motion, forces, equilibrium, and other advanced physics concepts.
WEEK 3 LESSON NOTE
Subject: Physics
Class: Senior Secondary School 2 (SS2)
Term: Second Term
Week: Three (3)
Topic: Equations of Uniformly Accelerated Motion
Duration: 40 minutes × 2 periods
SPECIFIC OBJECTIVES
By the end of this lesson, students should be able to:
Explain the meaning of uniformly accelerated motion.
Define velocity, acceleration, and time in relation to motion.
State the equations of uniformly accelerated motion.
Derive the equations of motion using simple logical steps.
Apply the equations of motion to solve numerical problems.
Distinguish between initial velocity and final velocity.
Solve simple real-life motion problems involving constant acceleration.
INSTRUCTIONAL MATERIALS
LESSON CONTENT
STEP 1: MEANING OF MOTION
Motion is said to occur when a body changes its position with time relative to a reference point.
Examples of motion include:
A car moving along a road
A ball rolling on the ground
A person walking from one place to another
To study motion properly, physicists consider quantities such as distance, displacement, velocity, acceleration, and time.
STEP 2: VELOCITY
Velocity is defined as the rate of change of displacement with time.
Velocity = Displacement / Time
Velocity is a vector quantity because it has both magnitude and direction.
Unit of velocity: metre per second (m/s)
There are two important types of velocity:
STEP 3: ACCELERATION
Acceleration is defined as the rate of change of velocity with time.
Acceleration = Change in velocity / Time taken
Change in velocity = Final velocity − Initial velocity
Therefore:
Acceleration (a) = (v − u) / t
Acceleration is also a vector quantity.
Unit of acceleration: metre per second squared (m/s²)
If velocity increases, acceleration is positive. If velocity decreases, acceleration is negative (this is called deceleration or retardation).
STEP 4: UNIFORMLY ACCELERATED MOTION
Uniformly accelerated motion is motion in which the velocity of a body changes at a constant rate.
This means that the acceleration remains constant throughout the motion.
Examples include:
STEP 5: EQUATIONS OF UNIFORMLY ACCELERATED MOTION
When a body moves with constant acceleration in a straight line, the following equations apply:
v = u + at
s = ut + (1/2)at²
v² = u² + 2as
Where:
u = initial velocity
v = final velocity
a = acceleration
t = time
s = displacement
These equations are known as the equations of motion.
STEP 6: DERIVATION OF THE FIRST EQUATION (v = u + at)
Acceleration is defined as:
Acceleration = (v − u) / t
Rearranging the equation:
v − u = at
Therefore:
v = u + at
STEP 7: DERIVATION OF THE SECOND EQUATION (s = ut + (1/2)at²)
Average velocity = (Initial velocity + Final velocity) / 2
Average velocity = (u + v) / 2
But displacement:
s = Average velocity × Time
Substituting v = u + at:
s = (u + (u + at)) / 2 × t
s = (2u + at) / 2 × t
s = ut + (1/2)at²
STEP 8: DERIVATION OF THE THIRD EQUATION (v² = u² + 2as)
From the first equation:
v = u + at
Making t the subject:
t = (v − u) / a
Substitute into the second equation:
s = ut + (1/2)at²
After substitution and simplification:
v² = u² + 2as
STEP 9: APPLICATIONS OF THE EQUATIONS OF MOTION
The equations of motion are used to:
Calculate stopping distances of vehicles
Determine speed of moving objects
Analyze free-fall motion
Solve motion problems in physics and engineering
STEP 10: SOLVED EXAMPLES
Example 1:
A car starts from rest and accelerates uniformly at 2 m/s² for 5 s. Find its final velocity.
Given:
u = 0 m/s
a = 2 m/s²
t = 5 s
Using v = u + at:
v = 0 + (2 × 5) = 10 m/s
Example 2:
A body moving with an initial velocity of 4 m/s accelerates uniformly at 3 m/s² for 6 s. Find the displacement.
Using s = ut + (1/2)at²:
s = (4 × 6) + (1/2 × 3 × 36)
s = 24 + 54 = 78 m
EVALUATION
Define uniformly accelerated motion.
State the three equations of motion.
Distinguish between initial velocity and final velocity.
A body accelerates uniformly from rest at 5 m/s² for 4 s. Calculate its final velocity.
Explain the physical meaning of acceleration.
CLASS WORK
Write out the three equations of motion and explain the symbols used.
A car accelerates from rest at 4 m/s² for 10 s. Find the final velocity.
Calculate the displacement of a body moving with u = 2 m/s, a = 1 m/s², and t = 8 s.
HOME WORK
Define velocity and acceleration.
State the units of velocity and acceleration.
A stone is thrown vertically upward with an initial velocity of 20 m/s. Calculate:
(a) Its velocity after 2 s
(b) The displacement after 2 s
CONCLUSION
The lesson is concluded by stressing that the equations of uniformly accelerated motion form the foundation for understanding motion in physics. Students are encouraged to practice solving numerical problems regularly, as mastery of these equations is essential for topics such as projectiles, free fall, and simple harmonic motion.
WEEK 4 LESSON NOTE
Subject: Physics
Class: Senior Secondary School 2 (SS2)
Term: Second Term
Week: Four (4)
Topic: Projectiles
Duration: 40 minutes × 2 periods
SPECIFIC OBJECTIVES
By the end of this lesson, students should be able to:
Explain the meaning of projectile motion.
Identify examples of projectiles in everyday life.
Describe the path followed by a projectile.
Resolve velocity into horizontal and vertical components.
State and explain the assumptions made in projectile motion.
Apply equations of motion to solve problems involving projectiles.
Calculate time of flight, maximum height, and horizontal range of a projectile.
INSTRUCTIONAL MATERIALS
LESSON CONTENT
STEP 1: MEANING OF PROJECTILE MOTION
Projectile motion refers to the motion of an object that is projected into the air and then moves under the influence of gravity alone.
Once a body is projected, no force acts on it except the force of gravity (air resistance is neglected).
A body undergoing projectile motion is called a projectile.
STEP 2: EXAMPLES OF PROJECTILES
Common examples of projectiles include:
A stone thrown into the air
A football kicked at an angle
A bullet fired from a gun
A javelin thrown by an athlete
In all these cases, the object follows a curved path.
STEP 3: PATH OF A PROJECTILE
The path followed by a projectile is called its trajectory.
In projectile motion, the trajectory is a parabola.
This parabolic path results from the combination of:
STEP 4: ASSUMPTIONS MADE IN PROJECTILE MOTION
For simplicity, the following assumptions are made:
Air resistance is negligible.
Acceleration due to gravity is constant.
The Earth is flat over the range of motion.
The projectile is small compared to the Earth.
These assumptions help make calculations easier and sufficiently accurate for basic physics.
STEP 5: RESOLUTION OF VELOCITY
When a body is projected at an angle to the horizontal, its initial velocity can be resolved into two perpendicular components:
If the projectile is launched with velocity u at an angle θ to the horizontal:
Horizontal component, ux = u cos θ
Vertical component, uy = u sin θ
These components act independently of each other.
STEP 6: HORIZONTAL MOTION OF A PROJECTILE
In the horizontal direction:
Horizontal distance covered:
x = ux × t
Where t is the time of flight.
STEP 7: VERTICAL MOTION OF A PROJECTILE
In the vertical direction:
The equations of uniformly accelerated motion apply:
v = u − gt
s = ut − (1/2)gt²
v² = u² − 2gs
STEP 8: TIME OF FLIGHT
Time of flight is the total time the projectile remains in the air.
Time of flight, T = (2u sin θ) / g
STEP 9: MAXIMUM HEIGHT
Maximum height is the greatest vertical distance reached by the projectile.
Maximum height, H = (u² sin² θ) / (2g)
At maximum height, the vertical velocity is zero.
STEP 10: HORIZONTAL RANGE
Horizontal range is the horizontal distance covered by the projectile before landing.
Range, R = (u² sin 2θ) / g
The range is maximum when θ = 45°.
STEP 11: SOLVED EXAMPLE
A ball is projected with a velocity of 20 m/s at an angle of 30° to the horizontal. Take g = 10 m/s². Calculate:
(a) Time of flight
(b) Maximum height
(c) Horizontal range
Given:
u = 20 m/s
θ = 30°
g = 10 m/s²
(a) Time of flight:
T = (2 × 20 × sin 30°) / 10
T = (40 × 0.5) / 10 = 2 s
(b) Maximum height:
H = (20² × sin² 30°) / (2 × 10)
H = (400 × 0.25) / 20 = 5 m
(c) Horizontal range:
R = (20² × sin 60°) / 10
R = (400 × 0.866) / 10 = 34.64 m
EVALUATION
Define projectile motion.
State four assumptions made in projectile motion.
Write expressions for the horizontal and vertical components of velocity.
State the formula for time of flight.
Explain why the path of a projectile is parabolic.
CLASS WORK
A stone is thrown with a speed of 15 m/s at an angle of 45°. Calculate its time of flight.
State two examples of projectile motion in sports.
Explain the meaning of trajectory.
HOME WORK
Define projectile and projectile motion.
A ball is projected with a velocity of 25 m/s at an angle of 60°. Calculate its maximum height.
Explain why air resistance is neglected in basic projectile motion.
CONCLUSION
The lesson is concluded by emphasizing that projectile motion combines horizontal uniform motion and vertical uniformly accelerated motion. Students are reminded that understanding projectiles is essential for topics such as ballistics, sports physics, and later studies in mechanics.
WEEK 5 LESSON NOTE
Subject: Physics
Class: Senior Secondary School 2 (SS2)
Term: Second Term
Week: Five (5)
Topic: Introduction to Simple Harmonic Motion (SHM)
Duration: 40 minutes × 2 periods
SPECIFIC OBJECTIVES
By the end of this lesson, students should be able to:
Explain oscillatory motion and simple harmonic motion.
Identify examples of simple harmonic motion in everyday life.
Define key terms used in SHM such as amplitude, period, frequency, and equilibrium position.
State the conditions necessary for a body to execute simple harmonic motion.
Explain displacement, velocity, and acceleration in SHM.
Relate restoring force to displacement in SHM.
Distinguish between oscillatory motion and simple harmonic motion.
INSTRUCTIONAL MATERIALS
LESSON CONTENT
STEP 1: MEANING OF OSCILLATORY MOTION
Oscillatory motion is a type of motion in which a body moves repeatedly to and fro about a fixed point or position.
In oscillatory motion, the body follows the same path repeatedly and the motion is periodic in nature.
Examples of oscillatory motion include:
Motion of a simple pendulum
Vibration of a tuning fork
Motion of a mass attached to a spring
STEP 2: MEANING OF SIMPLE HARMONIC MOTION
Simple Harmonic Motion (SHM) is a special type of oscillatory motion in which the restoring force acting on the body is directly proportional to the displacement from the equilibrium position and always acts towards that position.
Mathematically:
Restoring force ∝ displacement
Or
F = −kx
Where:
STEP 3: CONDITIONS FOR SIMPLE HARMONIC MOTION
For a body to execute simple harmonic motion, the following conditions must be satisfied:
The body must be capable of oscillating about a fixed point.
There must be a restoring force acting on the body.
The restoring force must be directly proportional to the displacement from the equilibrium position.
The restoring force must always act towards the equilibrium position.
STEP 4: EQUILIBRIUM POSITION
The equilibrium position is the position where the net force acting on the body is zero.
At this position:
In SHM, the body moves back and forth through the equilibrium position.
STEP 5: AMPLITUDE
Amplitude is the maximum displacement of the body from the equilibrium position.
It is usually represented by the letter A.
Amplitude indicates the extent of oscillation of the body.
Unit of amplitude: metre (m)
STEP 6: PERIOD AND FREQUENCY
Period (T) is the time taken for one complete oscillation.
Unit of period: second (s)
Frequency (f) is the number of complete oscillations per second.
Frequency = 1 / Period
Or
f = 1 / T
Unit of frequency: hertz (Hz)
STEP 7: DISPLACEMENT IN SHM
Displacement in SHM refers to the distance of the oscillating body from the equilibrium position at any given time.
Displacement can be:
depending on the direction of motion.
STEP 8: VELOCITY IN SHM
The velocity of a body executing SHM is not constant.
This variation occurs because the restoring force changes with displacement.
STEP 9: ACCELERATION IN SHM
Acceleration in SHM is caused by the restoring force.
Acceleration is:
Acceleration is always directed towards the equilibrium position.
STEP 10: DIFFERENCE BETWEEN OSCILLATORY MOTION AND SHM
| Oscillatory Motion | Simple Harmonic Motion |
|---|
| Motion to and fro about a point | Special type of oscillatory motion |
| Restoring force may not be proportional to displacement | Restoring force proportional to displacement |
| Path may not be sinusoidal | Path is sinusoidal |
STEP 11: REAL-LIFE APPLICATIONS OF SHM
Motion of a pendulum clock
Vibrations of strings in musical instruments
Movement of suspension springs in vehicles
Alternating current generation
EVALUATION
Define oscillatory motion.
Define simple harmonic motion.
State three conditions necessary for SHM.
Explain the meaning of amplitude.
Distinguish between frequency and period.
CLASS WORK
Give three examples of simple harmonic motion.
Explain why the restoring force in SHM is always directed towards the equilibrium position.
Define equilibrium position.
HOME WORK
Define amplitude, period, and frequency.
State two differences between oscillatory motion and SHM.
Explain how velocity varies in simple harmonic motion.
CONCLUSION
The lesson is concluded by emphasizing that simple harmonic motion is a fundamental type of motion in physics. Understanding SHM helps students to explain vibrations and waves, which are important concepts in sound, light, and many physical systems.
WEEK 6 LESSON NOTE
Subject: Physics
Class: Senior Secondary School 2 (SS2)
Term: Second Term
Week: Six (6)
Topic: Applications of Simple Harmonic Motion (Pendulum, Springs, etc.)
Duration: 40 minutes × 2 periods
SPECIFIC OBJECTIVES
By the end of this lesson, students should be able to:
Explain how simple harmonic motion applies to real physical systems.
Describe the motion of a simple pendulum.
State and explain the factors affecting the period of a simple pendulum.
Apply the formula for the period of a simple pendulum.
Explain the motion of a mass–spring system.
State Hooke’s law and apply it to spring systems.
State and apply the formula for the period of a mass–spring system.
Identify everyday applications of SHM.
INSTRUCTIONAL MATERIALS
LESSON CONTENT
STEP 1: APPLICATIONS OF SIMPLE HARMONIC MOTION
Simple harmonic motion is not just a theoretical concept; it appears in many physical systems in everyday life and technology.
Some common systems that execute SHM include:
Simple pendulum
Mass attached to a spring
Vibrating strings and tuning forks
Oscillations in mechanical and electrical systems
In this lesson, emphasis is placed on the simple pendulum and the mass–spring system.
STEP 2: SIMPLE PENDULUM
A simple pendulum consists of a small heavy object called a bob suspended from a fixed point by a light, inextensible string.
When the bob is displaced slightly from its equilibrium position and released, it oscillates to and fro about the equilibrium position.
For small angular displacements, the motion of a simple pendulum is simple harmonic motion.
STEP 3: PERIOD OF A SIMPLE PENDULUM
The period (T) of a simple pendulum is the time taken to complete one full oscillation.
For small oscillations, the period of a simple pendulum is given by:
T = 2π √(l / g)
Where:
STEP 4: FACTORS AFFECTING THE PERIOD OF A SIMPLE PENDULUM
The period of a simple pendulum depends on:
Length of the pendulum
Acceleration due to gravity
The period does not depend on:
Increasing the length increases the period, while increasing g decreases the period.
STEP 5: APPLICATIONS OF THE SIMPLE PENDULUM
STEP 6: MASS–SPRING SYSTEM
When a mass is attached to a spring and displaced from its equilibrium position, it oscillates to and fro.
If the restoring force of the spring is proportional to the displacement, the motion is simple harmonic motion.
STEP 7: HOOKE’S LAW
Hooke’s law states that the extension of a spring is directly proportional to the applied force, provided the elastic limit is not exceeded.
Mathematically:
F = kx
Where:
STEP 8: PERIOD OF A MASS–SPRING SYSTEM
The period of oscillation of a mass–spring system is given by:
T = 2π √(m / k)
Where:
STEP 9: COMPARISON OF PENDULUM AND SPRING SYSTEMS
| Simple Pendulum | Mass–Spring System |
|---|
| Depends on length | Depends on mass |
| Restoring force due to gravity | Restoring force due to spring |
| Used in clocks | Used in shock absorbers |
STEP 10: REAL-LIFE APPLICATIONS OF SHM
EVALUATION
Define a simple pendulum.
State the formula for the period of a simple pendulum.
Mention two factors affecting the period of a pendulum.
State Hooke’s law.
Write the formula for the period of a mass–spring system.
CLASS WORK
A pendulum has a length of 1 m. Calculate its period. (Take g = 10 m/s²)
Define spring constant.
Mention two applications of the mass–spring system.
HOME WORK
Explain why the period of a pendulum does not depend on the mass of the bob.
A mass of 0.5 kg is attached to a spring of constant 200 N/m. Calculate the period of oscillation.
List four everyday applications of SHM.
CONCLUSION
The lesson is concluded by emphasizing that simple harmonic motion plays an important role in many physical systems. Understanding the behavior of pendulums and springs helps students appreciate how SHM is applied in clocks, vehicles, buildings, and many technological devices.
WEEK 7 LESSON NOTE
Subject: Physics
Class: Senior Secondary School 2 (SS2)
Term: Second Term
Week: Seven (7)
Topic: Force and Equilibrium of Particles
Duration: 40 minutes × 2 periods
SPECIFIC OBJECTIVES
By the end of this lesson, students should be able to:
Define force and state its effects.
Explain the concept of a particle in physics.
Define equilibrium of a particle.
Identify different types of forces acting on a particle.
State and apply the conditions for equilibrium of a particle.
Resolve forces into horizontal and vertical components.
Solve simple numerical problems involving equilibrium of particles.
INSTRUCTIONAL MATERIALS
Physics textbook
Graph sheets
Ruler
Meter rule
Charts showing force diagrams
Small objects (stone, book, ring)
LESSON CONTENT
STEP 1: MEANING OF FORCE
A force is a push or a pull that can change the shape, size, speed, or direction of motion of a body.
Forces are responsible for:
Starting motion
Stopping motion
Changing speed
Changing direction
Deforming objects
Unit of force: newton (N)
STEP 2: TYPES OF FORCES
Forces acting on a body can be classified into:
(a) Contact Forces
These are forces that act only when bodies are in physical contact.
Examples include:
Frictional force
Tension in a string
Normal reaction
(b) Non-Contact Forces
These forces act without physical contact between bodies.
Examples include:
Gravitational force
Magnetic force
Electrostatic force
STEP 3: MEANING OF A PARTICLE
In physics, a particle is a body whose size and shape can be neglected when considering its motion or equilibrium.
A particle is treated as a point mass, especially when the dimensions of the body do not affect the analysis.
Examples:
STEP 4: EQUILIBRIUM OF A PARTICLE
A particle is said to be in equilibrium when all the forces acting on it balance such that there is no resultant force.
When a particle is in equilibrium:
STEP 5: CONDITIONS FOR EQUILIBRIUM OF A PARTICLE
For a particle to be in equilibrium, the vector sum of all forces acting on it must be zero.
This gives two conditions in a plane:
Sum of horizontal forces = 0
Sum of vertical forces = 0
These conditions are written as:
ΣFx = 0
ΣFy = 0
STEP 6: RESOLUTION OF FORCES
Resolution of forces involves splitting a force into two perpendicular components, usually horizontal and vertical.
If a force F acts at an angle θ to the horizontal:
Horizontal component, Fx = F cos θ
Vertical component, Fy = F sin θ
Resolution of forces helps simplify equilibrium problems.
STEP 7: FREE BODY DIAGRAM
A free body diagram is a diagram that shows all the forces acting on a particle.
Steps in drawing a free body diagram:
Isolate the particle.
Represent the particle as a point.
Draw all forces acting on it with arrows.
Label each force clearly.
Free body diagrams are essential in solving equilibrium problems.
STEP 8: SOLVED EXAMPLE
A weight of 10 N is suspended by two strings making angles of 30° and 60° with the horizontal. Find the tensions in the strings.
Let the tensions be T1 and T2.
Resolving horizontally:
T1 cos 30° = T2 cos 60°
Resolving vertically:
T1 sin 30° + T2 sin 60° = 10
Solving these equations gives the values of T1 and T2.
EVALUATION
Define force.
State two effects of force.
Explain the meaning of equilibrium of a particle.
Write the conditions for equilibrium of a particle.
Resolve a force of 20 N acting at 30° to the horizontal.
CLASS WORK
Define a particle in physics.
Draw a free body diagram of a book resting on a table.
State two examples each of contact and non-contact forces.
HOME WORK
Explain why a particle in equilibrium can still be in motion.
A force of 15 N acts at an angle of 45° to the horizontal. Find its horizontal and vertical components.
State three real-life situations where forces are in equilibrium.
CONCLUSION
The lesson is concluded by emphasizing that understanding force and equilibrium of particles is essential for studying mechanics. Students are reminded that equilibrium principles are widely applied in engineering, construction, and everyday physical systems.
WEEK 8 LESSON NOTE
Subject: Physics
Class: Senior Secondary School 2 (SS2)
Term: Second Term
Week: Eight (8)
Topic: Equilibrium of Rigid Bodies & Moments of Forces
Duration: 40 minutes × 2 periods
SPECIFIC OBJECTIVES
By the end of this lesson, students should be able to:
Define a rigid body in physics.
Explain the conditions for equilibrium of a rigid body.
Define the moment of a force.
State the principle of moments.
Solve simple problems involving moments and torques.
Explain the difference between a particle and a rigid body in equilibrium.
Apply moments of forces to everyday examples.
INSTRUCTIONAL MATERIALS
Physics textbook
Graph sheets
Ruler
Meter rule
Charts showing rigid bodies and lever systems
Small rods, weights, and pivot setups
LESSON CONTENT
STEP 1: DEFINITION OF RIGID BODY
A rigid body is a body that does not change its shape or size when forces are applied.
Unlike a particle, a rigid body has dimensions, and forces can produce rotation as well as translation.
Examples:
STEP 2: EQUILIBRIUM OF A RIGID BODY
A rigid body is in equilibrium when it is at rest or moves with constant velocity and the sum of forces and sum of moments about any point is zero.
Conditions for equilibrium:
Translational equilibrium: The vector sum of all forces acting on the body is zero.
Rotational equilibrium: The sum of clockwise moments equals the sum of anticlockwise moments about any pivot.
STEP 3: MOMENT OF A FORCE
The moment of a force about a point (or pivot) is a measure of the turning effect of the force.
Moment = Force × Perpendicular distance from pivot
M = F × d
Where:
Unit: Newton metre (N·m)
STEP 4: PRINCIPLE OF MOMENTS
The principle of moments states:
For a body in equilibrium, the sum of clockwise moments about a pivot equals the sum of anticlockwise moments about the same pivot.
ΣM_clockwise = ΣM_anticlockwise
STEP 5: LEVERS AND APPLICATIONS
A lever is a rigid rod that can rotate about a fixed pivot (fulcrum).
The effort, load, and fulcrum are the main components.
Types of levers:
First-class lever: Fulcrum between effort and load (e.g., seesaw, crowbar)
Second-class lever: Load between effort and fulcrum (e.g., wheelbarrow)
Third-class lever: Effort between fulcrum and load (e.g., hockey stick)
The mechanical advantage is achieved by applying the principle of moments.
STEP 6: SOLVED EXAMPLES
Example 1:
A 50 N weight is placed 2 m from a pivot. How much effort is needed 4 m from the pivot on the opposite side to lift it?
Using principle of moments:
Effort × distance = Load × distance
E × 4 = 50 × 2
E = 100 / 4 = 25 N
Example 2:
A uniform rod 3 m long weighs 60 N and is pivoted at one end. A force of 30 N is applied at the other end. Determine if it is in equilibrium.
Moments about pivot:
Clockwise moment = 60 × (3/2) = 90 N·m
Anticlockwise moment = 30 × 3 = 90 N·m
Since ΣM_clockwise = ΣM_anticlockwise, the rod is in equilibrium.
EVALUATION
Define a rigid body.
State the two conditions for equilibrium of a rigid body.
Write the formula for the moment of a force.
Explain the principle of moments.
Solve a simple problem using the principle of moments.
CLASS WORK
Define rotational equilibrium.
A lever has effort of 20 N applied 2 m from pivot and load of 30 N placed 1 m from pivot. Determine if the lever is balanced.
Draw a labelled diagram of a first-class lever.
HOME WORK
A uniform beam of weight 80 N is 4 m long. It is pivoted at one end. Find the anticlockwise moment of a 50 N force applied at 3 m from pivot.
Explain why the sum of moments is zero for a body in equilibrium.
List three real-life examples of the use of levers.
CONCLUSION
The lesson is concluded by emphasizing that understanding the equilibrium of rigid bodies and moments of forces is crucial in engineering, construction, and mechanics. The principle of moments allows students to calculate forces and design stable structures efficiently.
WEEK 9 LESSON NOTE
Subject: Physics
Class: Senior Secondary School 2 (SS2)
Term: Second Term
Week: Nine (9)
Topic: Center of Gravity, Stability & Equilibrium in Fluids (Archimedes’ Principle, Flotation)
Duration: 40 minutes × 2 periods
SPECIFIC OBJECTIVES
By the end of this lesson, students should be able to:
Define center of gravity (CG).
Determine the center of gravity of simple objects.
Explain stability of bodies and factors affecting stability.
Define fluid and state the properties of fluids.
Explain Archimedes’ Principle.
Solve problems involving buoyant force.
Explain the principle of flotation and its applications.
INSTRUCTIONAL MATERIALS
Physics textbook
Graph sheets
Ruler
Meter rule
Objects of different shapes (rod, flat sheet, cardboard)
Spring balance
Container with water
Small solid objects (wood, metal, plastic)
LESSON CONTENT
STEP 1: CENTER OF GRAVITY (CG)
The center of gravity of a body is the point at which the entire weight of the body may be considered to act.
For uniform objects, the CG is at the geometric center.
Examples:
Determining CG experimentally:
Suspend the object from a point and let it hang freely.
Draw a vertical line along the string.
Repeat from another point.
Intersection of lines gives the CG.
STEP 2: STABILITY OF BODIES
A body is said to be stable if it returns to its original position after being slightly displaced.
Types of equilibrium:
Stable equilibrium: CG rises when displaced slightly, body returns to original position.
Unstable equilibrium: CG falls when displaced, body topples.
Neutral equilibrium: CG remains at the same height, body stays in new position.
Factors affecting stability:
STEP 3: FLUIDS
A fluid is a substance that can flow and take the shape of its container.
Properties of fluids:
Examples: liquids, gases
STEP 4: ARCHIMEDES’ PRINCIPLE
Archimedes’ Principle states:
"A body wholly or partially immersed in a fluid experiences an upward force equal to the weight of the fluid displaced."
This upward force is called buoyant force (F_b).
Mathematically:
F_b = ρ × V × g
Where:
ρ = density of fluid (kg/m³)
V = volume of fluid displaced (m³)
g = acceleration due to gravity (m/s²)
STEP 5: BUOYANCY AND FLOATATION
Density and flotation:
STEP 6: REAL-LIFE APPLICATIONS
Ships and boats floating on water
Submarines controlling buoyancy
Hydrometers measuring liquid density
Balloons rising in air
STEP 7: SOLVED EXAMPLE
Example:
A block of wood of volume 0.02 m³ and density 600 kg/m³ is placed in water (density 1000 kg/m³). Find the upthrust and determine if it will float.
Weight of water displaced = ρ × V × g
= 1000 × 0.02 × 10
= 200 N
Weight of block = mass × g = (ρ × V) × g = 600 × 0.02 × 10 = 120 N
Since weight of block < upthrust, the block will float.
EVALUATION
Define center of gravity.
State three types of equilibrium.
Explain Archimedes’ principle.
Define buoyant force.
Solve a simple flotation problem.
CLASS WORK
Determine the CG of a uniform cardboard sheet using suspension method.
Explain why a ship does not sink even though it is heavy.
Define stability of a body.
HOME WORK
Explain factors affecting stability of a body.
A metal cube of volume 0.001 m³ and density 8000 kg/m³ is placed in water. Calculate the buoyant force and determine if it will float.
List four real-life applications of Archimedes’ principle.
CONCLUSION
The lesson is concluded by emphasizing that the concepts of center of gravity, stability, and equilibrium in fluids are crucial in engineering, shipbuilding, and everyday physics. Students should understand how buoyancy and stability determine whether objects float, sink, or topple.
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